91Ó°ÊÓ

1. Mass of sphere 1 is 50 g = 0.05 kg
2. Height of sphere 1 is 9.0 cm
3. Mass of sphere 2 is 85 g = 0.085 kg
4. Hint: Do not use rounded-off values.

Use the law of conservation of energy to find the lowest point velocity of the first ball. Using this velocity and conservation of momentum, find the velocity after each collision. Using the law of conservation of energy again, find the height achieved by each ball after each collision.

Formulae are as follow:

$$\begin{gathered} \frac{1}{2}m{v}^2=mgh \\ {v}_{1f}=\frac{m_1-m_2}{m_1+m_2}×v_{1i} \\ {v}_{2f}=\frac{2m}{m_1+m_2}×v_{1i} \end{gathered}$$

Here, m, m₁, m₂ are masses, v is velocity, g is an acceleration due to gravity and h is height.

Consider the diagram for the condition:

Here, use conservation of energy first to find initial velocity as follows:

Let’s assume,$v_{i1}$is the velocity of the first ball before the collision with the second ball,

$$\begin{gathered} mg{h}_1=\frac{1}{2}m{v}_{1i}^2 \\ {v}_{1i}=\sqrt{2gh} \\ {v}_{1i}=\sqrt{2×9.81×0.09} \\ {v}_{1i}=1.3288\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, velocity after collision of first ball, is$v_{1f}$can be found using equation 9-67 from the book as follows,

$v_{1f}=\frac{m_1-m_2}{m_1+m_2}×v_{1i}$

Negative sign because sphere 1 moves to left,

$$\begin{gathered} v_{1f}=\frac{0.05-0.085}{0.05+0.085}×1.3288 \\ {v}_{1f}=0.34450\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

After collision height $h_1$of sphere 1 can be found using law of conservation of energ

$$\begin{gathered} m_{1}g{h}_1=\frac{1}{2}{m}_1{v}_{1f}^2 \\ {h}_1=\frac{v_{1f}^2}{2g} \\ {h}_1=\frac{{(-0.34450)}^2}{2×9.81} \\ {h}_1=0.060489\mathrm{m} \end{gathered}$$

Hence, height reached by sphere 1 is 0.060489m.

Now, find the velocity of the second sphere after collision using equation 9-68 from the book as,

$$\begin{gathered} v_{2f}=\frac{2m_1}{m_1+m_2}×v_{1i} \\ {v}_{2f}=\frac{2×0.05}{0.05+0.085}×1.3288 \\ {v}_{2f}=0.98429\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Height of sphere 2 is, according to conservation of energy.

$$\begin{gathered} h_2=\frac{v_{2f}^2}{2g} \\ {h}_2=\frac{0.{985}^2}{2×9.8} \\ {h}_2=0.04943\mathrm{m} \end{gathered}$$

Hence, height reached by sphere 2 is 0.04943 m.

For the nextcollision height, sphere 1 would move towards right and sphere 2 would move in the left hand direction. Now, find the velocities as,

At lowest point sphere 1 has velocity,$v_{ff1}=\sqrt{2g{h}_1}$

$$\begin{gathered} v_{ff1}=\sqrt{2×9.8×0.006} \\ {v}_{ff1}=0.34432\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

At lowest point sphere 2 has velocity,$v_{ff2}=-\sqrt{2g{h}_2}$

$$\begin{gathered} v_{ff2}=\sqrt{2×9.8×0.049} \\ {v}_{ff2}=0.398429\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Now, after collision velocity for sphere 1 using equation 9.75 from the book, write$v_{ff}$as,

$$\begin{gathered} v_{ff}=\frac{m_1-m_2}{m_1+m_2}{v}_{ff1}+\frac{2m_1}{m_1+m_2}{v}_{ff2} \\ {v}_{ff}=\frac{0.05-0.085}{0.05+0.085}0.34432+\frac{2×0.085}{0.05+0.085}(-0.98429) \\ {v}_{ff}=-1.3287\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Now, according to energy conservation,

$$\begin{gathered} m_{1}g{h}_{ff1}=0.5×m_1×v_{ff} \\ {h}_{ff1}=\frac{v_{ff}^2}{2g} \\ {h}_{ff1}=\frac{{(-1.3287)}^2}{(2×9.8)} \\ {h}_{ff1}=0.09007\mathrm{m} \end{gathered}$$

This is the same as the original height of sphere 1.

Hence, after collision height reached by sphere 1 is 0.09007 m.

From part (c) for the sphere 1, after second collision, the height is obtained as, whichis the sameas the original height. It means all the energy of the system is now the potential energy of sphere 1. From the law of conservation of energy, if sphere 1 has all the energy,sphere2 must have zero energy. That implies, the positionof sphere2is the sameas its initial position.

Hence,after collision height reached by sphere 2 is zero.

Therefore, the law of conservation of energy and the equation for the elastic collision can be used to find the height of the spheres.